The category $bf{FinVect}$ of finite vector spaces is rigid.












4












$begingroup$


I am following Pavel Etingof et al's book on tensor categories.



They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".



My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.










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$endgroup$








  • 1




    $begingroup$
    It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
    $endgroup$
    – BWW
    Dec 4 '18 at 11:33










  • $begingroup$
    It's the diagonal matrices.
    $endgroup$
    – Kevin Carlson
    Dec 5 '18 at 2:14
















4












$begingroup$


I am following Pavel Etingof et al's book on tensor categories.



They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".



My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
    $endgroup$
    – BWW
    Dec 4 '18 at 11:33










  • $begingroup$
    It's the diagonal matrices.
    $endgroup$
    – Kevin Carlson
    Dec 5 '18 at 2:14














4












4








4





$begingroup$


I am following Pavel Etingof et al's book on tensor categories.



They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".



My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.










share|cite|improve this question











$endgroup$




I am following Pavel Etingof et al's book on tensor categories.



They give FinVect as an example of a rigid monoidal category, with evaluation map given by $text {ev}_V(epsilonotimes v)=epsilon(v)$ and coevaluation map $text{coev}_V:mathbb K rightarrow Votimes V^*$ "the usual embedding".



My question is embarassingly simple: what is this "usual embedding?" I was trying to see what it had to be using the definition of evaluation and coevaluation, but got nowhere.







category-theory monoidal-categories






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 13:33







Soap

















asked Dec 4 '18 at 11:27









SoapSoap

1,022615




1,022615








  • 1




    $begingroup$
    It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
    $endgroup$
    – BWW
    Dec 4 '18 at 11:33










  • $begingroup$
    It's the diagonal matrices.
    $endgroup$
    – Kevin Carlson
    Dec 5 '18 at 2:14














  • 1




    $begingroup$
    It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
    $endgroup$
    – BWW
    Dec 4 '18 at 11:33










  • $begingroup$
    It's the diagonal matrices.
    $endgroup$
    – Kevin Carlson
    Dec 5 '18 at 2:14








1




1




$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33




$begingroup$
It's touched on a bit here: en.wikipedia.org/wiki/Tensor_product#Relation_to_dual_space
$endgroup$
– BWW
Dec 4 '18 at 11:33












$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14




$begingroup$
It's the diagonal matrices.
$endgroup$
– Kevin Carlson
Dec 5 '18 at 2:14










1 Answer
1






active

oldest

votes


















3












$begingroup$

What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$



where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.



This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.



Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$






share|cite|improve this answer









$endgroup$









  • 6




    $begingroup$
    To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
    $endgroup$
    – Aaron
    Dec 4 '18 at 11:52













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









3












$begingroup$

What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$



where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.



This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.



Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$






share|cite|improve this answer









$endgroup$









  • 6




    $begingroup$
    To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
    $endgroup$
    – Aaron
    Dec 4 '18 at 11:52


















3












$begingroup$

What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$



where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.



This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.



Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$






share|cite|improve this answer









$endgroup$









  • 6




    $begingroup$
    To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
    $endgroup$
    – Aaron
    Dec 4 '18 at 11:52
















3












3








3





$begingroup$

What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$



where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.



This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.



Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$






share|cite|improve this answer









$endgroup$



What Etingof et al call the "usual embedding" is given by linear extension of the map$$1_Kmapstosum_{i = 1}^{n}v_iotimes v_i^* $$



where the $v_i$ are a basis of V and $v_i^*$ denoted the basis dual to the $v_i$'s.



This makes perfect sense if you want $ev circ fotimes idcirc coev = Trace(f)$ for any $f:Vlongrightarrow V$.



Try $f=id$ and you'll find that $$evcirc coev = dim(V)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 11:44









NephryNephry

514512




514512








  • 6




    $begingroup$
    To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
    $endgroup$
    – Aaron
    Dec 4 '18 at 11:52
















  • 6




    $begingroup$
    To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
    $endgroup$
    – Aaron
    Dec 4 '18 at 11:52










6




6




$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52






$begingroup$
To rephrase things slightly, if $V$ is finite dimensional, there is an isomorphism with $hom(V,V)$ and $Votimes V^{*}$, and the usual embedding is the inclusion of scalar maps into all linear maps.
$endgroup$
– Aaron
Dec 4 '18 at 11:52




















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